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Given a permutation $\sigma\in S_{n}$, is there a way to know the order of the centraliser $C_{S_{n}}\left(\sigma\right)=\left\{ \pi\in S_{n},\,\pi\sigma=\sigma\pi\right\}$ , i.e what is $\left|C_{S_{n}}\left(\sigma\right)\right|$?

I would appreciate a proof if the answer is yes.

Also, if the answer above is yes, is there also a way to calcualte the order of the centraliser of a given subset of $S_{n}$, or at least for a pair of permutations?

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3 Answers

Let $n_1,n_2,\ldots,n_k$ be the distinct lengths of the cycles of $\sigma$ (including 1 if there are fixed points) and suppose that there are $m_i$ cycles of length $n_i$. Then the centralizer of $\sigma$ can permute the cycles of the same length. Its order is $\prod_{i=1}^k n_i^{m_i}m_i!$.

Calculating the centralizer of a subgroup $H$ of $S_n$ is not difficult, but it is more complicated. The order of the centralizer of a single orbit is equal to the number of fixed points (in that orbit) of the stabilizer of a point in the orbit. But if $H$ has more than one orbit with equivalent actions then the equivalent orbits can be permuted by the centralizer, so the complete centralizer is a direct product of wreath products of centralizers of sets of equivalent orbits.

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i must say i didn't exactly understand the second part. any chance of getting an example with 2 different permutation or something of that sort? Thanks in any case! –  IBS Nov 28 '11 at 13:28
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You let the permutations act by conjugation on the permutation and you seek the size of the stabilizer of $\sigma$. By the orbit-stabilizer theorem, it is enough to know the size of the orbit, which is the well-known size of the conjugacy class of $\sigma$.

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Factor $\sigma$ into its orbits, and let $m_1, m_2, \ldots, m_i$ be the order of each of those orbits, with $m = \sum_{j=1}^i m_j$. For a permutation $\pi$ to commute with $\sigma$, for each of the orbits of $\pi$ we must have either that it is a power of an orbit of $\sigma$, or that it acts on none of the elements that $\sigma$ does. Thus we have that the centralizer group is isomorphic to

$$ \mathbb{Z}_{m_1}\oplus \mathbb{Z}_{m_2}\oplus \cdots \oplus\mathbb{Z}_{m_i} \oplus S_{n-m} $$

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This is wrong, and I'm sorry for writing something like this thinking it was an actual answer. The point that I've found that I'm wrong is that even though a set of orbits in $\pi$ and in $\sigma$ aren't powers of one another, they might both be powers of some other permutation. Example: in $S_6$, you have the translation $\phi = (1, 2, 3, 4, 5, 6)$. Let $\sigma = \phi^2 = (1, 3, 5)(2, 4, 6)$ and $\pi = \phi^3 = (1, 4)(2, 5)(3, 6)$. Then $\sigma$ and $\pi$ commutes without any of their respective orbits commuting. Disregard my answer, please. –  Arthur Nov 27 '11 at 6:09
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